1. soving exponential equations #2

$\displaystyle 2^ {x^2} \;=\;32(2^{4x})$

$\displaystyle 2^ {x^2} \;=\;32(2^{4x})$
$\displaystyle 2^ {x^2} \;=\;32(2^{4x})\Leftrightarrow 2^{x^2}=2^{5+4x}\implies x^2=5+4x$

3. but...

Originally Posted by felper
$\displaystyle 2^ {x^2} \;=\;32(2^{4x})\Leftrightarrow 2^{x^2}=2^{5+4x}\implies x^2=5+4x$
thanks, but how do you solve for x?